Question: The lifespans of turtles in a particular zoo are normally distributed. The average turtle lives $97$ years; the standard deviation is $7.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a turtle living less than $111.8$ years.
Solution: $97$ $89.6$ $104.4$ $82.2$ $111.8$ $74.8$ $119.2$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $97$ years. We know the standard deviation is $7.4$ years, so one standard deviation below the mean is $89.6$ years and one standard deviation above the mean is $104.4$ years. Two standard deviations below the mean is $82.2$ years and two standard deviations above the mean is $111.8$ years. Three standard deviations below the mean is $74.8$ years and three standard deviations above the mean is $119.2$ years. We are interested in the probability of a turtle living less than $111.8$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the turtles will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the turtles will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $82.2$ years and the other half $({2.5\%})$ will live longer than $111.8$ years. The probability of a particular turtle living less than $111.8$ years is ${95\%} + {2.5\%}$, or $97.5\%$.